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Suppose, I have to prove that $A\equiv B$.

I started out by proving that $¬B \implies ¬A$. This proves $A\implies B$. Next I proved that suppose B is true and A is not and this turns out to be contradiction. So $B\implies A$. I have some feeling that I might have proved same thing thing twice. However, mathematically these seem to be sound.

Is this mathematically sound?

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Nop, you're done, though I wouldn't express that $\,\neg B\Longrightarrow \neg A\,$ "proves" $\,A\Longrightarrow B\,$ , but rather I'd say that it is (logically) equivalent to it. –  DonAntonio Feb 13 '13 at 2:27
    
thank you for the remark. –  user45099 Feb 13 '13 at 2:39

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up vote 2 down vote accepted

Perfectly fine, (not knowing what $A$, $B$ are, or the details of your proof).

That said, your strategy is fine: to prove equivalence you need to prove $A \iff B \equiv (A\rightarrow B) \land (B\rightarrow A)$, which it looks like you have done, because $\lnot B \rightarrow \lnot A \equiv A\rightarrow B$, so having proved $\lnot B \rightarrow \lnot A$, you've established, equivalently, $A\rightarrow B$.

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A side question. If I have to prove that $A \implies B$ and I want to prove $A$ but $¬B$ leads to contradiction, is it sufficient to show that out of all possible cases, a single one leads to contradiction or do I have to show that all lead to contradiction? e.g. A=natural number. B=integer greater than 0. I want to prove $A \implies B$. Is saying that suppose $a$ satisfies $A$ but if it does not satisfy $B$, then it cannot satisfy $A$ or do I have to prove for every single cases? If it's unclear do let me know. –  user45099 Feb 13 '13 at 9:42

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